10 2 Engineering Analysis with Numerical Solutions (p 340) Some of the practical applications of integration are presented in Section 2 3 in Chapter 2 (p 38)
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10 2 Engineering Analysis with Numerical Solutions (p 340) Some of the practical applications of integration are presented in Section 2 3 in Chapter 2 (p 38)
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Chapter 10
Numerical Solution Methods for Engineering Analysis (Chapter 10 Numerical solution methods)© Tai-Ran HsuBased on the textbook on "Applied Engineering
Analysis", by Tai-Ran Hsu, published by John
Wiley & Sons, 2018 (ISBN 97811119071204)
Applied Engineering Analysis
- slides for class teaching* 1 2Chapter Learning Objectives
(p.339) perform integrations, and solve differential equations. performing integrations, and solving differential equations by the Runge-Kutta methods. solutions to the problems that are not readily or possibly solved by closed-form solution methods. solution points, and the accuracy of the solution is largely depending on the size of the increments of the variable selected for the solutions. packages such as Mathematica and MatLAB. 3 10.1Introduction
Numerical methods are techniques by which the mathematical problems involved with the engineering analysis cannot readily or possibly be solved by analytical methods such as those presented in previous chapters of this book. We will learn from this chapter on the use of some of these numerical methods that will not only enable engineers to solve many mathematical problems, but they will alsoallow engineers to minimizethe needs for the many hypotheses and idealization of the conditions, as stipulated in Section 1.4 (p.8) for engineering analysis.
This chapter will cover the principles of commonly used numerical techniques for: (1) the solution of nonlinear polynomial and transcendental equations, (2) Integration with integrals that involve complex forms of functions, and (3) the solution of differential equations by selected finite difference methods, (4) overviews of two popular commercial software packages called Mathematica andMatLAB.
4 10.2Engineering Analysis with Numerical Solutions
(p.340) There are a number of unique characteristics of numerical solution methods in engineeringanalysis. Following are just a few obvious ones:1) Numerical solutions are available only at selected (discrete) solution points, but not at all points
covered by the functions as in the case with analytical solution methods.2) Numerical methods are essentially "trail-and-error" processes. Typically, users need to
estimate an initial solution with selected increment of the variable to which the intended solution will cover. Unstable numerical solutions may result from improper selection of step sizes (the incremental steps) with solutions either in the form of "wild oscillation" or becoming unbounded in the trend of values.3) Most numerical solution methods results in errors in the solutions. There are two types of errors that
are inherent with numerical solutions:(a) Truncation errors - Because of the approximate nature of numerical solutions, they often consists
of lower order terms and higher order terms. The latter terms are often dropped in the computations for the sake of computational efficiency, resulting in error in the solution, and (b) Round-off errors -Most digital computers handle either numbers with 7 decimal points, or 14decimal points in numerical solutions. In the case of 32-bit computer with double precision (i.e. 14
decimal points length numbers), any number after the 14 th decimal point will be dropped. This may not sound like a big deal, but if a huge number of operations are involved in the computation, such error can accumulate and result in significant error in the end results. Both these errors are of accumulative natures. Consequently, errors in numerical solution may grow to be significant with solutions obtained after many step with the set increments. 5 10.3Solution of Nonlinear Equations
(p.341) We have learned the distinction between linear and nonlinear algebraic equations in Section 4.1. There are numerous occasions that engineers are requested to solve nonlinear equations such as the equation for the solution t f of the following nonlinear equation in Example 8.9 on page 270:We reported a solution of t
f =0.7 in Equation (10.2) by a "short cut" solution method, and also t f = 0.862 bya more accurate solution method such as the Newton- Raphson method described in Section 10.3.2. (10.2)
There are a number of numerical methods available to solve nonlinear equations such as in Equation (10.2);
what we will introduce here in the book are the following two methods that are readily available by using
digital computers:10.3.1
Solution using Microsoft Excel software (Example 10.1) (p.342): In this method, we will first express the equation in the form of f(x)=0 as shown in Figure 10.1. For example, we will express Equation (10.3) in Example 10.1 from the form of x 4 -2x 3 +x 2 -3x=-3 into the form: x 4 -2x 3 +x 2 -3x+3=0, in which we will get the function f(x) = x 4 -2x 3 +x 2 -3x+3.The roots "x" would lie in the range between x=x
i and x i+1 with which the values of f(x i ) and f(x i+1 ) bearing different positive or negative sign. The difference of (x i-1 ) and (x i or between x i+1 and x i is referred to be the increment of x-value, or is expressed as ǻx.Figure 10.1 Roots in Nonlinear Equation
f(x) = 0 610.3.1
Solution using Microsoft Excel software (Example 10.1) -Cont'd
Example 10.1
Solve the nonlinear polynomial Equation in (10.3): x 4 -2x 3 +x 2 -3x+3=0Solution:
We have Equation 10.3 expressed as f(x)=0 with f(x) = x 4 -2x 3 +x 2 -3x+3. We will use Microsoft Excel software to evaluate the function f(x) with an increment of the variable x, ǻx=0.5 beginning at x =0. The values of the function f(x) with x i (i = 1,2,3,....,9) are shown in the Table in the right, and the plot of function f(x) vs. variable x is depicted in Figure 10.2 ix f(x)1 0 3.00
2 0.5 1.56
31.0 0
41.5 -0.94
52.0 +1.00
6 2.5 9.56
7 3.0 30.00
8 3.5 69/06
9 4.0 135.00
We notice from the computed values of f(x) with variable x in Figure 10.2 that there are two roots of the
equation in the ranges of (x=1.0 and 1.5) and the other root in the range of (x = 1.5 and 2.0) because the
sign changes of the function f(x) cross these two ranges of x variable. The first root of x =1 is obvious
because it resulted in f(x) = 0. The search of the second root with computations of the function f(x) with
smaller increment of x between x = 1.5 and x=2.0 indicated an approximate root at x = 1.8 as illustrated in
the plot of the results in Figure 10.2.